Oct 19 Curvature

Simple Definition of Curvature

Before we start formally defining curvature, we must at least have a basic idea of what it should be. So curvature is how curvy something is. What I mean is, if something doesn't curve at all, like y = x, the curvature is 0. If something is extremely curvy, then it's curvature is really high. Now, I know that's very hand-wavy, so let's get to some real math.

Definitions

So, we need to define a few things.

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The unit tangent vector is defined as:
$$T(t) = \dfrac{r'(t)}{|r'(t)|}$$
Because $|T(t)|$ is constant (it's always 1), by a theorem we proved before (which isn't on this site, sorry), $T(t)$ is perpendicular to $T'(t)$. So, now we have our unit normal vector.
The unit normal vector is defined as:
$$N(t) = \dfrac{T'(t)}{|T'(t)|}$$
This is always perpendicular to the normal vector.

Another way

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Sometimes, it's not good to differentiate with t, after all we don't want to define curvature based on time. If something goes across a curve at 2t speed, that would change the curvature. So, let's define it based on arclength, denoted s.
$$s = \int_0^{s}\mathrm{\sqrt{f'(t)^2 + g'(t)^2}}\, \mathrm{d}t$$
If we take the derivative with respect to s on both sides, we get
$$1 = \sqrt{(f'(s))^2 + (g'(s))^2}$$
Note that arclength also equals the magnitude of r'(s), so $$|r'(s)| = 1$$ That makes it equivalent to the unit tangent vector, so we call it $T(s)$.

Curvature

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Let $\theta$ equal the angle between the unit tangent vector and $\hat{i}$. So we can define curvature, denoted k as
$$K = |\dfrac{d\theta}{ds}|$$
Now, we can use this to derive another formula for curvature. If C is the graph y = f(x), then the curvature is
$$K = \dfrac{|y''|}{(1 + (y')^2)^\frac{3}{2}}$$
Let's say we have x = f(t), and y = g(t). Curvature then equals
$$K = \dfrac{|f'(t)g''(t) - g'(t)f''(t)|}{(f'(t)^2 + g'(t)^2)^{3/2}}$$
You can prove that by dividing $\frac{d\theta}{dt}$ over $\frac{ds}{dt}$

3D Curvature Formula

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If x = f(t), y = g(t), z = h(t),
$$K = \dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}$$
Note: this is similar to the 2D formula. The denominator is equal to the $\sqrt{f'(t)^2 + g'(t)^2}^3$, which is equal to the denominator of the 2D curvature formula. The numerator is just like the 2D version, just 3D.